Graduate How to solve this integral? (something to do with a beta distribution?)

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SUMMARY

The integral discussed is expressed in Maple notation as Int(exp(c[0]*ln(y)/a[0]+c[1]*M*ln(M-y)/a[1]), y = 0 .. M); and simplifies to a beta function. The transformation y = Mt leads to the integral M^{(c[0]/a[0]) + (Mc[1]/a[1]) + 1} ∫_0^1 t^{c[0]/a[0]} (1 - t)^{c[1]/a[1]} dt. When substituting values a[1] = 0.7, a[0] = 0.4, c[1] = 0.9, c[0] = 0.5, and M = 7, the results are 17939559.61 and 344821202.1 for the respective integrals.

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TL;DR
Solving the integral (in Maple notation): Int(exp(c[0]*ln(y)/a[0]+c[1]*M*ln(M-y)/a[1]), y = 0 .. M);
I have the following integral (in Maple notation):

Int(exp(c[0]*ln(y)/a[0]+c[1]*M*ln(M-y)/a[1]), y = 0 .. M);

with (in Maple notation):

0<a[0], 0<a[1], 0<c[0], 0<c[1], 0<y, y<M, 0<M.

What is the solution of this integral? I suspect that the solution has something to do with a beta distribution.
 
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Simplifying the integrand and substituting y = Mt yields <br /> \int_0^M \exp\left(<br /> c_0 \frac{\ln(y)}{a_0} + c_1 M \frac{\ln(M - y)}{a_1}<br /> \right)\,dy
$$= \int_0^M y^{c_0/a_0}(M - y)^{Mc_1/a_1}\,dy =
M^{(c_0/a_0) + (Mc_1/a_1) + 1} \int_0^1 t^{c_0/a_0} (1 - t)^{c_1/a_1}\,dt
$$ which is indeed a beta function.
 
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Thank you very much. You were of great help.
 
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If I enter the values ##a_{1}\, = \, 0.7##, ##a_{0}\, = \, 0.4##, ##c_{1}\, = \, 0.9##, ##c_{0}\, = \, 0.5## and ##M\, = \,7## in
$$\int_{0}^{M}\!{y}^{{\frac {c_{0}}{a_{0}}}} \left( M-y \right) ^{{\frac {Mc_{1}}{a_{1}}}}\,{\rm d}y$$
I get ## 17939559.61##. If I enter the same values in
$${M}^{{\frac {c_{0}}{a_{0}}}+{\frac {Mc_{1}}{a_{1}}}+1}\int_{0}^{1}\! \left( 1-x \right) ^{{\frac {c_{1}}{a_{1}}}}{x}^{{\frac {c_{0}}{a_{0}}}}\,{\rm d}x$$
I get ##344821202.1##.
 
That should be <br /> \int_0^M y^{c_0/a_0}(M - y)^{Mc_1/a_1}\,dy =<br /> M^{(c_0/a_0) + (Mc_1/a_1) + 1} \int_0^1 t^{c_0/a_0} (1 - t)^{Mc_1/a_1}\,dt
 
Again thank you very much. Sorry for my mistake.
 
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